Compiler-Design
September 10, 2024Computer-Networks
September 11, 2024Probability
| Question 1 |
Let A, B and C be independent events which occur with probabilities 0.8, 0.5 and 0.3 respectively. The probability of occurrence of at least one of the event is __________
| 0.93 |
Question 1 Explanation:
P(A∪B∪C) = P(A) + P(B) + P(C) – P(A∩B) – P(B∩C) – P(A∩C) + P(A∩B∩C)
Since all the events are independent, so we can write
P(A∪B∪C) = P(A) + P(B) + P(C) – P(A)P(B) – P(B)P(C) – P(A)P(C) + P(A)P(B) P(C)
= 0.8 + 0.5 + 0.3 – 0.4 – 0.5 – 0.24 + 0.12
= 0.93
Since all the events are independent, so we can write
P(A∪B∪C) = P(A) + P(B) + P(C) – P(A)P(B) – P(B)P(C) – P(A)P(C) + P(A)P(B) P(C)
= 0.8 + 0.5 + 0.3 – 0.4 – 0.5 – 0.24 + 0.12
= 0.93
Correct Answer: A
Question 1 Explanation:
P(A∪B∪C) = P(A) + P(B) + P(C) – P(A∩B) – P(B∩C) – P(A∩C) + P(A∩B∩C)
Since all the events are independent, so we can write
P(A∪B∪C) = P(A) + P(B) + P(C) – P(A)P(B) – P(B)P(C) – P(A)P(C) + P(A)P(B) P(C)
= 0.8 + 0.5 + 0.3 – 0.4 – 0.5 – 0.24 + 0.12
= 0.93
Since all the events are independent, so we can write
P(A∪B∪C) = P(A) + P(B) + P(C) – P(A)P(B) – P(B)P(C) – P(A)P(C) + P(A)P(B) P(C)
= 0.8 + 0.5 + 0.3 – 0.4 – 0.5 – 0.24 + 0.12
= 0.93
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